Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y^2 + 6y - 27}{y - 3} \times \dfrac{y - 1}{-3y - 27} $
Explanation: First factor the quadratic. $z = \dfrac{(y + 9)(y - 3)}{y - 3} \times \dfrac{y - 1}{-3y - 27} $ Then factor out any other terms. $z = \dfrac{(y + 9)(y - 3)}{y - 3} \times \dfrac{y - 1}{-3(y + 9)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (y + 9)(y - 3) \times (y - 1) } { (y - 3) \times -3(y + 9) } $ $z = \dfrac{ (y + 9)(y - 3)(y - 1)}{ -3(y - 3)(y + 9)} $ Notice that $(y - 3)$ and $(y + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ \cancel{(y + 9)}(y - 3)(y - 1)}{ -3(y - 3)\cancel{(y + 9)}} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $z = \dfrac{ \cancel{(y + 9)}\cancel{(y - 3)}(y - 1)}{ -3\cancel{(y - 3)}\cancel{(y + 9)}} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $z = \dfrac{y - 1}{-3} $ $z = \dfrac{-(y - 1)}{3} ; \space y \neq -9 ; \space y \neq 3 $